Two moles of an ideal gas are allowed to expand from a volume of $10$ $d m^{3}$ to $2$ $m^{3}$ at $300$ K against a pressure of $101.325$ KPa. Calculate the work done?

- 201.6

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13.22

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- 810.6

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- 18.96

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Solution

The correct option is A $- 201.6$ kJ
$Δ V = V_{2} - V_{1} = 2 m^{3} - (10 d m^{3} \times 10^{- 3} m^{3} / d m^{3}) = 1.99 m^{3}$
$P = 101.325 k P a = 101.325 \times 10^{3} P a$
$W = - P Δ V$
$W = - 101.325 \times 10^{3} P a \times 1.99 m^{3}$
$W = - 201.6 \times 10^{3} J$
$W = - 201.6 k J$

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Two moles of an ideal gas are allowed to expand from a volume of 10 dm3 to 2 m3 at 300K against a pressure of 101.325 KPa. Calculate the work done?

The correct option is A −201.6 kJΔV=V2−V1=2 m3−(10 dm3×10−3 m3/dm3)=1.99 m3P=101.325 kPa=101.325×103 PaW=−PΔVW=−101.325×103 Pa×1.99 m3W=−201.6×103 JW=−201.6 kJ

Two moles of an ideal gas are allowed to expand from a volume of $10$ $d m^{3}$ to $2$ $m^{3}$ at $300$ K against a pressure of $101.325$ KPa. Calculate the work done?

The correct option is A $- 201.6$ kJ
$Δ V = V_{2} - V_{1} = 2 m^{3} - (10 d m^{3} \times 10^{- 3} m^{3} / d m^{3}) = 1.99 m^{3}$
$P = 101.325 k P a = 101.325 \times 10^{3} P a$
$W = - P Δ V$
$W = - 101.325 \times 10^{3} P a \times 1.99 m^{3}$
$W = - 201.6 \times 10^{3} J$
$W = - 201.6 k J$